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3 years ago

Juan’s game board is a square with sides that are 50 cm long. The area of the large circle is about 1,963.5 square cm.

Mathematics

2 answers:

Mashcka [7]3 years ago

8 0

Answer:

C) 0.785

Step-by-step explanation:

I am assuming that hitting the blue area will result in 0 points.

The area of said square is 50*50, which is 2500 cm².

The chances of scoring >0 points is the chance of hitting the circle.

That is 1963.5/2500.

Therefore, the chance is 0.7854, or 0.785.

Alex Ar [27]3 years ago

4 0

Answer:

Im trying to think in mt pee-wee brain about how to do this..

Step-by-step explanation:

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If c is the hypotenuse of a right triangle,find the missing side. If necessary, round to the nearest hundredth. c=sqrt(155), b=s

Lunna [17]

A^2 + 31 = 155 - I squared each of b and c for those numbers
a^2 = 155 - 31
a^2 = 124
a = 4 * sqrt(31)

6 0

3 years ago

Find the probability of at least 6 failures in 7 trials of a binomial experiment in which the probability of success in any one

oksano4ka [1.4K]

Answer:

$P(x \geq 6)=P(X=6)+P(X=7)$

And we can find the individual probabilities:

$P(X=6)=(7C6)(0.91)^6 (1-0.91)^{7-6}=0.358$

$P(X=7)=(7C7)(0.91)^7 (1-0.91)^{7-7}=0.517$

And replacing we got:

$P(x \geq 6)=P(X=6)+P(X=7)= 0.358+0.517=0.875$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=1-0.09=0.91)$

The probability associated to a failure would be p =1-0.09 = 0.91

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

And we want to find this probability:

$P(x \geq 6)=P(X=6)+P(X=7)$

And we can find the individual probabilities:

$P(X=6)=(7C6)(0.91)^6 (1-0.91)^{7-6}=0.358$

$P(X=7)=(7C7)(0.91)^7 (1-0.91)^{7-7}=0.517$

And replacing we got:

$P(x \geq 6)=P(X=6)+P(X=7)= 0.358+0.517=0.875$

6 0

3 years ago

The profit that a vendor makes per day by selling x elephant ears is given by the function P(x)= -0.004x^2+3.2x-200. Find the nu

dezoksy [38]

Answer:

The number of elephant ears that must be sold to maximize profit is 400.

Step-by-step explanation:

Given that,

The profit that a vendor makes per day is given by

P(x)= - 0.004x² +3.2 x -200

where x is number of elephant ears.

P(x)= - 0.004x² +3.2 x -200

Differentiating with respect to x

P'(x)= - 0.008x+3.2

Again differentiating with respect to x

P''(x) = -0.008

For maximum or minimum P'(x)=0

- 0.008x+3.2=0

⇒0.008x=3.2

$\Rightarrow x=\frac{3.2}{0.008}$

⇒ x = 400

$P''(x)|_{x=400} = -0.008$

Since at x=400, P''(x)<0, the profit is maximize.

P(400) = -0.004×400²+3.2×400-200

=440

The number of elephant ears that must be sold to maximize profit is 400.

3 0

3 years ago

Solve y=f(x) for x. Then find the input when the output is -3.

DochEvi [55]

Answer:

Please check the explanation

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:\left(x-5\right)^3-1$

Given that the output = -3

i.e. y = -3

now substituting the value y=-3 and solve for x to determine the input 'x'

$\:\:y=\:\left(x-5\right)^3-1$

$-3\:=\:\left(x-5\right)^3-1\:\:\:$

switch sides

$\left(x-5\right)^3-1=-3$

Add 1 to both sides

$\left(x-5\right)^3-1+1=-3+1$

$\left(x-5\right)^3=-2$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

Thus, the input values are:

$x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}$

And the real input is:

$x=-\sqrt[3]{2}+5$

$x=3.74$

4 0

2 years ago

Please help me out I need it done ASAP

Anni [7]

Your answer will be letter b

3 0

3 years ago